Whether the switch $K$ is open or closed, the reading of galvanometer is the same. If $I$ denotes the current, then:

I_{1} R_{4} = I G

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I_{1} R_{5} = I G

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I_{1} R_{3} = I G

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I_{1} R_{5} = I R_{3}

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Solution

The correct option is B $I_{1} R_{5} = I G$
If the reading of galvanometer does not change, then Wheatstone bridge with $R_{3}, R_{4}, R_{5}, G$ and key $K$ is balanced.

$I_{1} (R_{4}) = I (R_{3})$
$I_{1} (R_{5}) = I G$

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